# How do you find the volume of the region below y= -3x+6 and enclosed by the y-axis from 0 to 2, rotated about the x-axis?

Jun 11, 2015

#### Answer:

${V}_{\text{tot}} = \frac{56}{9} \pi .$

#### Explanation:

First of all, we have to understand how the region of the plane is made:
The region is determined by this system:
$\left\{\begin{matrix}y \ge 0 \\ y \le 2 \\ y \le - 3 x + 6 \\ x \ge 0\end{matrix}\right.$
And the points interested are:
$P = \left\{A \left(0 , 2\right) , B \left(\frac{4}{3} , 2\right) , C \left(2 , 0\right) , D \left(0 , 0\right)\right\}$ This is the formula for calculating the volume of a solid obtained from a revolution around the x-axis: $V = \pi \cdot {\int}_{a}^{b} {f}^{2} \left(x\right) \mathrm{dx}$
But in this case f(x) is a composite function:
$f \left(x\right) = \left\{\begin{matrix}y = 2 \mathmr{if} 0 \le x \le \frac{4}{3} \\ y = - 3 x + 6 \mathmr{if} \frac{4}{3} \le x \le 2\end{matrix}\right.$

So we should break the solid into two rotation solids:
a cylinder from x=0 to x=4/3 and a cone from x=4/3 to x=2:
${V}_{\text{tot"=V_"cylinder"+V_"cone}}$
${V}_{\text{cylinder}} = \pi {\int}_{0}^{\frac{4}{3}} {\left(2\right)}^{2} \mathrm{dx} = \pi {\left[4 x\right]}_{0}^{\frac{4}{3}} = \frac{16}{3} \pi .$

${V}_{\text{cone}} = \pi {\int}_{\frac{4}{3}}^{2} {\left(- 3 x + 6\right)}^{2} \mathrm{dx} = \pi {\int}_{\frac{4}{3}}^{2} \left(9 {x}^{2} - 36 x + 36\right) \mathrm{dx}$

${V}_{\text{cone}} = \pi {\left[3 {x}^{3} - 18 {x}^{2} + 36 x\right]}_{\frac{4}{3}}^{2} = \pi \left(3 \cdot {2}^{3} - 18 \cdot {2}^{2} + 36 \cdot 2 - 3 {\left(\frac{4}{3}\right)}^{3} + 18 {\left(\frac{4}{3}\right)}^{2} - 36 \left(\frac{4}{3}\right)\right)$

${V}_{\text{cone}} = \pi \left(24 - 72 + 72 - \frac{64}{9} + \frac{288}{9} - 48\right) = \pi \left(- \frac{216}{9} - \frac{64}{9} + \frac{288}{9}\right) = \frac{8}{9} \pi$

${V}_{\text{tot}} = \frac{16}{3} \pi + \frac{8}{9} \pi = \frac{56}{9} \pi .$